What is inverse oflaplace (s+5)/(s^2+6s+18) ?

The answer to inverse oflaplace (s+5)/(s^2+6s+18) is e^{-3t}cos(3t)+2/3 e^{-3t}sin(3t)

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inverse oflaplace (s+5)/(s^2+6s+18)

inverse laplace

\frac{s + 5}{s ^{2} + 6 s + 18}

e^{- 3 t} cos (3 t) + \frac{2}{3} e^{- 3 t} sin (3 t)

Solution steps

L^{- 1} {\frac{s + 5}{s ^{2} + 6 s + 18}}

Expand

\frac{s + 5}{s ^{2} + 6 s + 18} : \frac{s + 3}{( s + 3 ) ^{2} + 9} + 2 \cdot \frac{1}{( s + 3 ) ^{2} + 9}

= L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 9} + 2 \cdot \frac{1}{( s + 3 ) ^{2} + 9}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 9}} + 2 L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 9}}

L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 9}} : e^{- 3 t} cos (3 t)

L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 9}} : e^{- 3 t} \frac{1}{3} sin (3 t)

= e^{- 3 t} cos (3 t) + 2 e^{- 3 t} \frac{1}{3} sin (3 t)

Refine

e^{- 3 t} cos (3 t) + 2 e^{- 3 t} \frac{1}{3} sin (3 t) : e^{- 3 t} cos (3 t) + \frac{2}{3} e^{- 3 t} sin (3 t)

= e^{- 3 t} cos (3 t) + \frac{2}{3} e^{- 3 t} sin (3 t)

y^{^{''}} - 6 y^{^{'}} + 9 y = 6 x^{2} + 2 - 12 e^{3 x}

(3 + sin (2 x))^{2}, at x = 0

\frac{2}{x ^{2} - x}

\int 2 x + 8 x^{- 2} d x

f (x) = x^{2} + 10 x, at x = 8

What is inverse oflaplace (s+5)/(s^2+6s+18) ?
The answer to inverse oflaplace (s+5)/(s^2+6s+18) is e^{-3t}cos(3t)+2/3 e^{-3t}sin(3t)